How Does the Substitution s=1-t Simplify the Integral in Calculating f(π/3)?

[Azael]

the problem is this
$$f(x)=\int_{1/2}^{sin^2x} \frac{ cos^2t }{ t^2-t }dt - \int_{1/2}^{cos^2x} \frac{sin^2(1-t]}{t^2-t}dt$$
solve $$f(\frac{\pi}{3})$$
They begin by doing the substitution s=1-t on the second integral and it leads to $$t^2-t=s^2-s$$ and $$dt=-ds$$
this leads to the 2 integrals
$$f(\frac{\pi}{3})=\int_{1/2}^{3/2} \frac{cos^2t}{t^2-t}dt + \int_{1/2}^{3/2} \frac{sin^2s}{s^2-s}ds$$
they then just make it into
$$f(\frac{\pi}{3})=\int_{1/2}^{3/2} \frac{1}{t^2-t}dt$$
Solving that one is a pieace of cake

but its this last step I don't follow. Why can I say that $$sin^2s+cos^2t=1$$? How can it be 1 when its 2 different variables??
and why is $$t^2-t=s^2-s$$ Why can I just replace the s with t's when s isn't equal to t?

 

[matt grime]

The answer to your question is that s and t are dummy variables.
$$\int f(s)ds +\int g(t)dt = \int f(s)ds +\int g(s)ds=\int (f(s)+g(s))ds$$

is a standard result and one you use all the time whether you realize it or not.

 

[Azael]

When I look at it as you wrote it it feels pretty obvious, thanks. I guess the whole part of defining s=1-t threw me off.